[LeetCode] 728. Self Dividing Numbers
阿新 • • 發佈:2018-08-04
return contains use == true left 不出 int ++
A self-dividing number is a number that is divisible by every digit it contains.
For example, 128 is a self-dividing number because
128 % 1 == 0
,128 % 2 == 0
, and128 % 8 == 0
.Also, a self-dividing number is not allowed to contain the digit zero.
Given a lower and upper number bound, output a list of every possible self dividing number, including the bounds if possible.
Example 1:
Input:
left = 1, right = 22
Output: [1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 15, 22]
Note:
- The boundaries of each input argument are 1 <= left <= right <= 10000.
打印出left
到right
範圍內的所有Self Dividing Numbers
1.可以被這個數包含的每個數字整除
2.數字中沒有0
拼拼湊湊寫出一個粗糙的版本,邏輯混亂,但我第一個版本的代碼就長這樣,不打草稿多思考就開寫就是這個後果
vector<int> selfDividingNumbers(int left, int right) { vector<int> result; for (int i = left; i <= right; i++) { if (i < 10) { result.push_back(i); continue; } if (i % 10 == 0) { continue; } bool isSelfDividing = true; int tmp = i; while (tmp % 10 != 0) { if (i % (tmp % 10) != 0) { isSelfDividing = false; } tmp /= 10; } if (tmp >= 10) { isSelfDividing = false; } if (isSelfDividing) { result.push_back(i); } } return result; }
這個版本的代碼太亂了,實在拿不出手,優化一下
1.小於10的數都是Self Dividing Numbers
2.不包含0(不能被10整除)
3.能被自身包含的每個數字整除(寫個for循環取模)
bool isSelfDividing(int num) { if (num < 10) { return true; } for (int i = num; i != 0; i /= 10) { if (i % 10 == 0) { return false; } if (num % (i%10) != 0) { return false; } } return true; } vector<int> selfDividingNumbers(int left, int right) { vector<int> result; for (int i = left; i <= right; i++) { if (isSelfDividing(i)) { result.push_back(i); } } return result; }
[LeetCode] 728. Self Dividing Numbers